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Question Number 74394 by aliesam last updated on 23/Nov/19

Commented by mathmax by abdo last updated on 23/Nov/19

$$1)letdecomposeF\left(x\right)=\frac{4x^2+x}{x^3-81x}\Rightarrow F\left(x\right)=\frac{4x^2+x}{x(x-9)(x+9)}$$$$=\frac{4x+1}{(x-9)(x+9)}=\frac{a}{x-9}+\frac{b}{x+9}$$$$a=(x-9)F\left(x\right){\mid }_{x=9}=\frac{37}{18}$$$$b=(x+9)F\left(x\right){\mid }_{x=-9}=\frac{-35}{-18}=\frac{35}{18}\Rightarrow F\left(x\right)=\frac{37}{18(x-9)}+\frac{35}{18(x+9)}\Rightarrow$$$$\int \frac{4x^2+x}{x^3-81x}dx=\int F\left(x\right)dx=\frac{37}{18}ln\mid x-9\mid +\frac{35}{18}ln\mid x+9\mid +C.$$

Commented by mathmax by abdo last updated on 23/Nov/19

$$letA=\int x^{3}ln(x+8)dx\Rightarrow A{=}_{x+8=t}\int {(t-8)}^{3}ln\left(t\right)dt$$$$=\int (t^3+3t^2.8+3t.8^2+8^3)ln\left(t\right)dt$$$$=\int t^{3}ln\left(t\right)dt+24\int t^{2}ln\left(t\right)dt+192\int tln\left(t\right)dt+512\int ln\left(t\right)dt$$$$\int lntdt=tlnt-t+c_1$$$$\int tln\left(t\right)dt{=}_{bypsrts}\frac{t^2}{2}lnt-\int \frac{t^2}{2}\frac{dt}{t}=\frac{t^{2}lnt}{2}-\frac{1}{2}\int tdt$$$$=\frac{t^{2}lnt}{2}-\frac{t^2}{4}+c_2$$$$\int t^{2}ln\left(t\right)dt=\frac{t^{3}ln\left(t\right)}{3}-\int \frac{t^3}{3}\frac{dt}{t}=\frac{t^{3}ln\left(t\right)}{3}-\frac{1}{3}\int t^{2}dt$$$$=\frac{t^{3}ln\left(t\right)}{3}-\frac{1}{9}{t}^3+c_3$$$$\int t^{3}ln\left(t\right)dt=\frac{t^{4}ln\left(t\right)}{4}-\int \frac{t^4}{4}\frac{dt}{t}=\frac{t^{4}ln\left(t\right)}{4}-\frac{1}{16}{t}^4+c_4\Rightarrow$$$$A=\frac{{(x+8)}^{4}ln(x+8)}{4}-\frac{1}{16}{(x+8)}^4+8{(x+8)}^{3}ln(x+8)-\frac{8}{3}{(x+8)}^3$$$$+192\{\frac{{(x+8)}^{2}ln(x+8)}{2}-\frac{{(x+8)}^2}{4}\}+512\{(x+8)ln(x+8)-(x+8)\}+C$$

Commented by ~blr237~ last updated on 23/Nov/19

$$f\left(x\right)=\int x^{3}ln(8+x)dx$$$$bypartwehave{u}^{\prime }=x^3\to u=\frac{1}{4}{x}^4-\frac{8^4}{4}$$$$andv=ln(8+x)\to v^{\prime }=\frac{1}{8+x}$$$$f\left(x\right)=\left[\left(\frac{x^4-8^4}{4}\right)ln(8+x)\right]-\frac{1}{4}{\int }^{}\frac{x^4-8^4}{x-(-8)}dx$$$$=\left(\frac{x^4-8^4}{4}\right)ln(8+x)-\frac{1}{4}\int \frac{8^4}{8}[1+(-\frac{x}{8})+{(-\frac{x}{8})}^2+{\left(\frac{-x}{8}\right)}^{3}dx$$$$=\left(\frac{x^4-8^4}{4}\right)ln(8+x)-4^5[x-\frac{x^2}{16}+\frac{x^3}{3\times 64}-\frac{x^4}{4\times 8^3}]+c$$$$So$$$$f\left(x\right)=\left(\frac{x^4-8^4}{4}\right)ln(8+x)-4^{5}x+4^3{x}^2-\frac{4^2{x}^3}{3}+\frac{x^4}{4}+c$$$$$$

Commented by MJS last updated on 24/Nov/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{one}$$

Answered by MJS last updated on 23/Nov/19

$$\ast$$$$\int \frac{4x^2+x}{x^3-81x}dx=\int \frac{4x+1}{x^2-81}dx=$$$$=\frac{37}{18}\int \frac{dx}{x-9}+\frac{35}{18}\int \frac{dx}{x+9}=$$$$=\frac{37}{18}\ln (x-9)+\frac{35}{18}\ln (x+9)+C$$$$$$$$\ast \ast$$$$\int x^3\ln (x+8)dx=$$$$[u^{\prime }=x^3\Rightarrow u=\frac{1}{4}{x}^4;v=\ln (x+8)\Rightarrow v^{\prime }=\frac{1}{x+8}]$$$$=\frac{1}{4}{x}^4\ln (x+8)-\frac{1}{4}\int \frac{x^4}{x+8}dx=$$$$$$$$\int \frac{x^4}{x+8}dx=\int (x^3-8x^2+64x-512+\frac{4096}{x+8})dx=$$$$=\frac{1}{4}{x}^4-\frac{8}{3}{x}^3+32x^2-512x+4096\ln (x+8)$$$$$$$$=-\frac{x(3x^3-32x^2+384x-6144)}{48}+(\frac{1}{4}{x}^4-1024)\ln (x+8)+C$$

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